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I'm attempting some questions from Zwiebach - A First Course in String Theory, and have got stuck. I've proved that a function $h'(u)$ is periodic. The question then asks me to show that $h(u)=au+f(u)$ where $a$ is a constant and $f(u)$ a periodic function. I can't see how to do this directly from the periodicity of $h'$. Is this possible, or true?

Many thanks!

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3 Answers 3

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We may assume without loss of generality that the period of $h'$ is $1$, so that $h'(u + 1) = h'(u)$.

Consider $h(u + 1) - h(u)$. By differentiation, we find that $h(u + 1) - h(u) = a$ for some constant $a$. One now guesses that $h(u) = a u + f(u)$ for some periodic function $f$. So, $$f(u + 1) - f(u) = h(u + 1) - h(u) - a = 0$$ hence $f$ is indeed periodic with period $1$.

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Hint: Set $a$ equal to the integral of $h'$ over one period divided by the length of one period.

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Since $h'$ is periodic, integrating $h'$ over an interval the size of its period will always give the same value, let's call it $I$. Integrating e.g. from $0$ to $T$ will give $I$, from $0$ to $2T$ will give $2I$, etc. In general, $\forall x$, $\int_x^{x+T} h'(u) du =I$. You'll see then that $h(x+T)-h(x)=I$. (take any point $x_0$ to define $h(x)=\int_{x_0}^x h' +h(x_0)$

Case 1: $I=0$. Then $h$ is a periodic function of period $T$.

Case 2: $I\neq 0$. Then consider the function $g=h'-I/T$. $g$ is also periodic of period $T$ and the integral of $g$ over it's period is $0$. We are back to case 1. You'll then find that $a=I/T$ and any integration constant can go in $f(u)$.

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